\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 52, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/52\hfil Existence of $\psi$-bounded solutions]
{Existence of $\psi$-bounded solutions for nonhomogeneous linear
differential equations}

\author[P. N. Boi\hfil EJDE-2007/52\hfilneg]
{Pham Ngoc Boi}

\address{Pham Ngoc Boi \newline
Department of Mathematics, Vinh University, Vinh City,
Vietnam}
\email{pnboi\_vn@yahoo.com}

\thanks{Submitted January 26, 2007. Published April 5, 2007.}
\subjclass[2000]{34A12, 34C11, 34D05}
\keywords{$\psi$-bounded solution; $\psi$-integrable;
 $\psi$-integrally bounded; \hfill\break\indent 
 $\psi$-exponential dichotomy}

\begin{abstract}
 In this article we present a necessary and sufficient condition
 for the existence of $\psi$-bounded solution on $\mathbb{R}$ of
 the nonhomogeneous linear differential equation $x'=A(t)x+f(t)$.
 We associate that with the condition of the concept $\psi$-dichotomy
 on $\mathbb{R}$ of the homogeneous linear differential equation
 $x'=A(t)x$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}

\section{Introduction}

The existence of $\psi$-bounded and $\psi$-stable solutions on
$\mathbb{R}_+$ for systems of ordinary differential equations has
been studied by many authors; see for example Akinyele \cite{a1},
Avramescu \cite{a2}, Constantin \cite{c1}, Diamandescu \cite{d1,d2,d3}.
 Denote by $\mathbb{R}^d$ the $d$-dimensional Euclidean space. Elements in
this space are denoted by $x=(x_1,x_2,\dots , x_d)^T$ and their
norm by $\|x\|=\max \{ |x_1|, |x_2|,\dots , | x_d|\}$. For real
$d\times d$ matrices, we define norm
$| A|= \sup_{\|x\|\leqslant 1}\|Ax\Vert$.
 Let $\mathbb{R}_+ =[0, \infty)$,
$\mathbb{R}_-=(- \infty,0]$,
$J=\mathbb{R}_-,\mathbb{R}_+$ or $\mathbb{R}$ and
$\psi_i :J \to (0,\infty)$, $i=1,2,\dots,d$ be continuous functions. Set
$$
\psi =\mathop{\rm diag}[\psi_1, \psi_2,\dots ,\psi_d].
$$

\begin{definition} \label{def1.1}  \rm
 A function $f:J \to \mathbb{R}^d$ is said to be
\begin{itemize}
\item $\psi$-bounded on $J$ if $\psi(t) f(t)$ is
bounded on $J$.
\item $\psi$-integrable on $J$ if $f(t)$ is measurable
and $\psi(t) f(t)$ is Lebesgue integrable on $J$.
\item $\psi$-integrally bounded on $J$ if $f(t)$ is measurable and
the Lebesgue integrals $\int_t^{t+1}\|\psi(u) f(u)\|du$ are
uniformly bounded for any $t,  t+1\in J$.
\end{itemize}
\end{definition}

 In $\mathbb{R}^d$, consider  the following equations
\begin{gather}
 x' = A(t)x+f(t),\label{e1.1} \\
 x' = A(t)x. \label{e1.2}
\end{gather}
where $A(t)$ is continuous matrix on $J$, $f(t)$ is a continuous function
on $J$. Let $Y(t)$ be fundamental matrix of  \eqref{e1.2} with
$Y(0)=I_d$, the identity $d\times d$ matrix.
 The $d\times d$ matrices $P_1, P_2$ is said to be the pair of the
supplementary projections if $P_1^2=P_1, P_2^2=P_2, P_1+P_2=I_d$.


\begin{definition} \label{def1.2} \rm
 The equation \eqref{e1.2} is said to have a $\psi$-exponential dichotomy
on $J$ if there exist  positive constants $K, L, \alpha, \beta$
and a pair of the supplementary projections $P_1, P_2 $ such that
\begin{gather}
|\psi(t) Y(t) P_1 Y^{-1}(s)\psi^{-1}(s) |\leqslant
 Ke^{-\alpha(t-s)} \quad\text{for }    s\leqslant   t, s,t\in J, \label{e1.3}
\\
|\psi(t) Y(t) P_2 Y^{-1}(s)\psi^{-1}(s)|\leqslant Le^{\beta(t-s)}
\quad\text{for }    t\leqslant   s, s,t \in J. \label{e1.4}
\end{gather}
The equation \eqref{e1.2} is said to have a $\psi$-ordinary
dichotomy on $J$ if \eqref{e1.3}, \eqref{e1.4} hold with
$\alpha =\beta =0$.

We say that  \eqref{e1.2} has a $\psi$-bounded grow if for some fixed $h>0$
there exists a constant $C\geqslant 1$ such that every solution $x(t)$
of \eqref{e1.2} is satisfied
\begin{equation}
\|\psi(t)x(t)\|\leqslant C\|\psi(s)x(s)\|\mbox{ for }
s\leqslant t\leqslant s+h, s,t\in J. \label{e1.5}
\end{equation}
\end{definition}

\begin{remark} \label{rmk1.3} \rm
It is easy to see that if  \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}_+$ and on $\mathbb{R}_-$ with a pair
of the supplementary projections $P_1, P_2$ then
\eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$
with the pair of the supplementary projections $P_1, P_2$.
\end{remark}

\begin{theorem}[\cite{b1,d1,d3}] \label{thm1.4}
(a) The equation \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R}_+$ for every  $\psi$-integrable function $f$ on
$\mathbb{R}_+$ if and only if  \eqref{e1.2} has a
$\psi$-ordinary dichotomy on $\mathbb{R}_+$.

\noindent (b)
The equation \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R}_+$ for every  $\psi$- integrally bounded function $f$ on
$\mathbb{R}_+$ if and only if  \eqref{e1.2} has a $\psi$-exponential dichotomy  on $\mathbb{R}_+$.

\noindent (c) Suppose that  \eqref{e1.2} has a $\psi$-bounded
grow on $\mathbb{R}_+$. Then,  \eqref{e1.1} has at least
one $\psi$-bounded solution on $\mathbb{R}_+$  for every $\psi$-bounded
function $f$ on $\mathbb{R}_+$  if and only if  \eqref{e1.2}
has a $\psi$-exponential dichotomy on $\mathbb{R}_+$.
\end{theorem}

 \begin{theorem}[\cite{d3}] \label{thm1.5}
 Suppose that  \eqref{e1.1} has a $\psi$-exponential dichotomy on
$\mathbb{R}_+$ and, $P_1\ne  0, P_2\ne 0$.
If $lim_{t\to\infty}\|\psi(t)f(t)\| =0$  then every $\psi$-bounded
solution $x(t)$ of  \eqref{e1.1} is such that
 $lim_{t\to\infty}\|\psi(t)x(t)\| =0$.
 \end{theorem}

\section{Preliminaries}

\begin{lemma} \label{lem2.1}
(a)  Let    \eqref{e1.2} has a $\psi$-exponential dichotomy on
 $\mathbb{R}_+$ with a pair of the supplementary projections $P_1, P_2$.
If $Q_1, Q_2$  is a pair of the supplementary projections such that
$ImP_1= ImQ_1$, then   \eqref{e1.2} also has a $\psi$-exponential
dichotomy on $\mathbb{R}_+$ with the pair of the supplementary projections
$Q_1, Q_2$.

\noindent (b) Let  \eqref{e1.2} have a $\psi$-exponential dichotomy
on $\mathbb{R}_-$ with a pair of the supplementary projections $P_1, P_2$.
If $Q_1, Q_2$  is a pair of supplementary projections such that
$ImP_2= ImQ_2$, then   \eqref{e1.2} also has a $\psi$-exponential
dichotomy on $\mathbb{R}_-$ with the pair of the supplementary projections
$Q_1, Q_2$.
\end{lemma}

\begin{proof}
First, we prove in the case of $J =\mathbb{R}_+$. Note that
\eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$ with
the pair of the supplementary projections $P_1, P_2$ if only if following
statements are satisfied:
\begin{gather}
 \|\psi(t)Y(t)P_1\xi\|\leqslant K'e^{-\alpha(t-s)}\|\psi(s)Y(s)\xi\|
\quad\mbox{for all }  \xi \in \mathbb{R}^d \mbox{ and }
 t \geqslant  s \geqslant 0 ,   \label{e2.1}\\
 \|\psi(t)Y(t)P_2\xi\|\leqslant L'e^{\beta(t-s)}\|\psi(s)Y(s)\xi\|
\quad \mbox{for all } \xi \in \mathbb{R}^d \mbox{ and }
 s \geqslant  t \geqslant 0.      \label{e2.2}
 \end{gather}
In fact, if \eqref{e1.3} and \eqref{e1.4} are true, we have for any
vector $y\in \mathbb{R}^d $
  \begin{gather*}
 \|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)y\|\leqslant Ke^{-\alpha(t-s)}\|y\|
 \quad \mbox{for } t \geqslant  s \geqslant 0 ,\\
 \|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)y\|\leqslant Le^{\beta(t-s)}\|y\|
 \quad \mbox{for } s\geqslant  t \geqslant 0.
 \end{gather*}
Choose $y=\psi(s)Y(s)\xi$, we obtain \eqref{e2.1}, \eqref{e2.2}.
Conversely, suppose that inequalities \eqref{e2.1}, \eqref{e2.2} are true.
For any vector $y\in \mathbb{R}^d $, putting $\xi=Y^{-1}(s)\psi^{-1}(s)y$
we get \eqref{e1.3}, \eqref{e1.4}.

Now prove the lemma. It follows from $Ker P_2= ImP_1= ImQ_1=KerQ_2$
that $P_2Q_1=0$. Hence $P_1Q_1= P_1Q_1+P_2Q_1=Q_1$. Similarly $Q_1P_1=P_1$.
Then
\begin{gather}
P_1-Q_1=P^2_1- P_1Q_1=  P_1(P_2-Q_2) \label{e2.3},\\
P_1-Q_1= - Q_1P_2 =P_1P_2 - Q_1P_2 =  (P_1-Q_1)P_2. \label{e2.4}
\end{gather}
For each $u\in \mathbb{R}^d $, put $\xi=(P_1-Q_1)u$. The relation
\eqref{e2.3} implies that $\xi\in ImP_1$, then $P_1\xi=\xi$.
Result from \eqref{e2.1}, for $s = 0$ that
\begin{equation} \label{e2.5}
\|\psi(t)Y(t)[P_1-Q_1]u\|\leqslant K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]u\|,
t\geqslant 0.
\end{equation}
By \eqref{e2.4} we conclude
\begin{equation} \label{e2.6}
\begin{aligned}
K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]u\|
&=K'e^{-\alpha t}\|\psi(0)[P_1-Q_1]P_2u\|\\
&\leqslant  K'|\psi(0)||P_1-Q_1| e^{-\alpha t}\|P_2u\|,\quad
t\geqslant 0.
\end{aligned}
\end{equation}
Applying \eqref{e2.2}, for $t=0$, we get
\begin{equation}
\begin{aligned}
\|P_2u\|&=\|\psi^{-1}(0)\psi(0)P_2u\|\\
&\leqslant |\psi^{-1}(0)|\|\psi(0)P_2u\|\\
&\leqslant L'e^{-\beta s}|\psi^{-1}(0)|\|\psi(s)Y(s)u\|, \quad \mbox{for}
 s \geqslant 0.
\end{aligned} \label{e2.7}
\end{equation}
The relations \eqref{e2.5}--\eqref{e2.7} imply
\begin{equation} \label{e2.8}
 \begin{aligned}
\|\psi(t)Y(t)[P_1-Q_1]u\|&\leqslant K'L'|\psi(0)||\psi^{-1}(0)|
|P_1-Q_1|e^{-\alpha t}e^{-\beta t}\|\psi(s)Y(s)u\| \\
& \leqslant K_1e^{\beta(t-s)}\|\psi(s)Y(s)u\|, \quad
\mbox{ for } t, s \geqslant  0 .
\end{aligned}
\end{equation}
On the other hand, by \eqref{e2.2} we get
\begin{equation} \label{e2.9}
\|\psi(t)Y(t)P_2u\|\leqslant L'e^{\beta(t-s)}\|\psi(sY(s))u\|
,\quad \mbox{ for } 0 \leqslant t\leqslant  s.
\end{equation}
It follows from $Q_2=P_2+P_1-Q_1,$ \eqref{e2.8} and \eqref{e2.9} that
\begin{equation} \label{e2.10}
\begin{aligned}
\|\psi(t)Y(t)Q_2u\|&\leqslant\|\psi(t)Y(t)P_2u\|+\|\psi(t)Y(t)[P_1-Q_1]u\|\\
&\leqslant (L'+K_1)e^{\beta(t-s)}\|\psi(s)Y(s)u\|\\
& \leqslant L_2e^{\beta(t-s)}\|\psi(s)Y(s)u\| ,\quad\mbox{for } 0
\leqslant t \leqslant  s.
\end{aligned}
\end{equation}
Similarly,  for $u \in\mathbb{ R}^d$, we have
\begin{equation} \label{e2.11}
\|\psi(t)Y(t)Q_1u\|\leqslant K_2e^{-\alpha(t-s)}\|\psi(s)Y(s)u\|
   ,\quad \mbox{for } 0 \leqslant  s \leqslant  t.
\end{equation}
Then from this inequality, \eqref{e2.10}  and the preceding note it follows
 that \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$
with the pair of the supplementary projections $Q_1, Q_2$.
In the case of  $J =\mathbb{ R}_ -$,  the proof is similar.
\end{proof}

\begin{remark} \label{rmk2.2} \rm
(a) Suppose that   \eqref{e1.2} has a $\psi$-exponential dichotomy
on $\mathbb{R}_+$ with a pair of supplementary projections $P_1, P_2$.
The set $P_1\mathbb{R}^d$  is the subspace of $\mathbb{R}^d$ consisting
of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on
$\mathbb{R}_+$ of   \eqref{e1.2}.
 In fact, denote by $X_1$ this subspace, if $v \in P_1 \mathbb{R}^d$
then $v \in X_1$ by virtue of (2.1). Conversely if $u \in X_1$,
 we have to show that $P_2u = 0$.
Suppose otherwise that $P_2u\ne 0$, by (2.1), (2.2) we have $\|\psi(t)Y(t)P_1u\|$
 is bounded and the limit of $\|\psi(t)Y(t)P_2u\|$   is $\infty$,
as $t$ tend to $\infty$. Denote $y$ the solution of  \eqref{e1.2}, $y(0) = u$.
The relation $\psi(t)y(t) -\psi(t)Y(t)P_1u = \psi(t)Y(t)P_2u$ follows that
$y$ is non$\psi$-bounded on $\mathbb{R}_ +$, which is a contradiction.

\noindent(b) Similarly if  \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}_-$ with a pair of supplementary projections
$P_1, P_2$ then the  set $P_2\mathbb{R}^d$  is the subspace of $\mathbb{R}^d$
consisting of the values $x(0)$ of all $\psi$-bounded solutions $x(t)$ on
$\mathbb{R}_-$ of   \eqref{e1.2}.

(c) Suppose that  \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}$, then  \eqref{e1.2} has no nontrivial
$\psi$-bounded solution on $\mathbb{R}$.
In fact if $x(t)$ is the $\psi$-bounded solution of \eqref{e1.2} on
$\mathbb{R}$ then it is $\psi$-bounded on $\mathbb{R}_+$ and on
$\mathbb{R}_-$. Because equation \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}_+$, and on $\mathbb{R}_-$ with a pair of
supplementary projections $P_1, P_2$,  by preceding notice we have
$P_2x(0) = 0$ and $P_1x(0) = 0$. Hence $x(0) = 0$, then $x(t)$ is the
trivial solution of  \eqref{e1.2}.
\end{remark}

\begin{lemma}[\cite{m1}] \label{lem2.3}
Let $h(t)$ be a non-negative, locally integrable
such that
  $$
\int_t^{t+1}h(s)ds\leqslant c, \quad \mbox{for all }  t \in\mathbb{ R}
$$
If $\theta > 0$ then, for all $t \in\mathbb{ R}$,
\begin{gather}
 \int_t^{\infty} e^{-\theta (s-t)}h(s)ds\leqslant c[1-e^{-\theta}]^{-1},
\label{e2.12}\\
\int_{-\infty}^t e^{-\theta (t-s)}h(s)ds\leqslant c[1-e^{-\theta}]^{-1}.
\label{e2.13}
\end{gather}
\end{lemma}

\begin{proof}
 We prove \eqref{e2.12}, the proof of  \eqref{e2.13} is similar.
\begin{align*}
\int_{t+m}^{t+m+1} e^{-\theta(s-t)}h(s)ds
&\leqslant  \int_{t+m}^{t+m+1} e^{-\theta(t+m)}e^{\theta t}h(s)ds \\
&=\int_{t+m}^{t+m+1} e^{-\theta m}h(s)ds\leqslant c e^{-\theta m}
\end{align*}
implies that
$$
\int_t^\infty e^{-\theta(s-t)}h(s)ds =\sum_{m=0}^\infty
 \int_{t+m}^{t+m+1}e^{-\theta(s-t)}h(s)ds\leqslant c\sum_{m=0}^\infty
e^{-\theta m} = c[1-e^{-\theta}]^{-1}
$$
\end{proof}

\begin{lemma} \label{lem2.4}
Equation \eqref{e1.1} has at least one $\psi$-bounded solution on
$\mathbb{R}$  for every $\psi$-integrally bounded  function $f$ on
$\mathbb{R}$  if and only if the following three conditions are satisfied:
\begin{enumerate}
\item Equation \eqref{e1.1} has at least one solution on $\mathbb{R}$,
$ \psi$-bounded on $\mathbb{R}_+$  for every $\psi$- integrally bounded
function $f$ on $\mathbb{R}_+$
\item  Equation \eqref{e1.1} has at least one solution on $\mathbb{R}$,
 $\psi$-bounded on $\mathbb{R}_-$ for every $\psi$-integrally bounded
function $f$ on $\mathbb{R}_-$.
\item  Every solution of  \eqref{e1.2} is the sum of
two solution of  \eqref{e1.2}, one of that is $\psi$-bounded
on $\mathbb{R}_+$, another is $\psi-$ bounded on $\mathbb{R}_-$.
\end{enumerate}
\end{lemma}

\begin{proof}
Suppose the three conditions are satisfied we have to prove that \eqref{e1.1}
has at least one $\psi$-bounded solution on $\mathbb{R}$ for every
$\psi$-integrally bounded  function $f$ on $\mathbb{R}$.
Every $\psi$-integrally bounded  function $f$ on $\mathbb{R}$ is
$\psi$-integrally bounded  function $f$ on $\mathbb{R}_+$ and on
$\mathbb{R}_-$. Then for each $\psi$-integrally bounded  function
$f$ on $\mathbb{R}$  exists the solution $y_1$ and $y_2$ of \eqref{e1.1},
which is defined on $\mathbb{R}$ and corresponding $\psi$-bounded on
$\mathbb{R}_+$ and on $\mathbb{R}_-$. Denote by $x(t)$ the solution
of  \eqref{e1.2} such that $x(0) = y_2(0)-y_1(0)$.
By 3, we get $x(t) = x_1(t)+x_2(t)$, here $x_1, x_2$ are two solutions
of  \eqref{e1.2}, that are corresponding $\psi$-bounded solution on
$\mathbb{R}_+$ and $\mathbb{R}_-$. Set $z_1 = y_1+x_1, z_2=y_2-x_2$.

Hence $z_1$ and $z_2$ are the solutions of  \eqref{e1.1} corresponding
$\psi$-bounded solution on $\mathbb{R}_+$ and on $\mathbb{R}_-$.
Further, $z_2(0)=y_2(0)-x_2(0)=y_1(0)+x_1(0)=z_1(0)$, then $z_1=z_2$.
Consequently $z_1$ is a $\psi$-bounded solution on $\mathbb{R}$
of  \eqref{e1.1}.

Conversely, now if  \eqref{e1.1} has at least one $\psi$-bounded solution
on $\mathbb{R }$ for every $\psi$-integrally bounded  function $f$
on $\mathbb{R}$ we have to prove three condition are satisfied.
The conditions 1, 2 are satisfied since every $\psi$-integrally bounded
 function $f$ on $\mathbb{R}_+$ , or $\mathbb{R}_- $ is the restriction
of a $\psi$- integrally bounded  function $f$ on $\mathbb{R}$.
We prove that the condition 3 is satisfied. Set
\[
h(t)=\begin{cases} 0& \mbox{for } | t|\geqslant 1
\\ 1 &\mbox{for } t=0\\
\mbox{linear} &\text{for } t\in[-1,0],t\in[0,1] \end{cases}
\]
 Fix a solution $x(t)$ of  \eqref{e1.2} . Then $h(t)x(t)$ is a
$\psi$-integrally bounded  function on $\mathbb{R}$.
Set $y(t) =x(t)\int_0^th(s)ds$ , we have
 $$
y'(t) = A(t)x(t)\int_0^th(s)ds +h(t)x(t)  = A(t)y(t)+h(t)x(t).
$$

By hypothesis, the equation
$$ y'(t)=A(t)y(t)+h(t)x(t)
$$
has a solution  $\widetilde y(t)$, which is $\psi$-bounded on $\mathbb{R}$.
Set $x_1(t) = \widetilde y(t)-y(t)+\frac{1}{2}x(t)$   and
$x_2(t)=  \widetilde y(t)+y(t)+\frac{1}{2}x(t)$. It follows from
$\int_{-1}^0h(t)dt =\int_0^1h(t)dt=\frac{1}{2}$  that
$x_1(t) = \widetilde y(t)$  for $t \geqslant 1$; $x_2(t) =\widetilde y(t)$
for $t\leqslant  -1$. Then $x_1, x_2$ are the corresponding
$\psi$-bounded solutions on $\mathbb{R}_+$, $ \mathbb{R}_ -$ of  \eqref{e1.2}.
Consequently the solution $x(t)$ of  \eqref{e1.2} is the sum of  two solutions
$x_1(t)$ and $ x_2(t)$ of  \eqref{e1.2}, those solutions satisfy the
condition 3. The lemma is proved.
\end{proof}

\section{Main results}

\begin{theorem} \label{thm3.1}
 Equation \eqref{e1.1} has at least one $\psi$-bounded solution on
$\mathbb{R}_-$  for every $\psi$-integrally bounded  function $f$
on $\mathbb{R}_-$ if and only if  \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}_-$.
\end{theorem}

\begin{proof}
 This Theorem can be shown as in \cite[Theorem 3.3]{b1}.
We give the main steps of the proof as follows.
In the proof of ``if part'': Suppose that
$\int_{t-1}^t\|\psi(s)f(s)\|ds\leqslant c$ for $t\leqslant 0$.
By using Lemma \ref{lem2.3} we get
\begin{align*}
 \|\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)ds\|
&\leqslant \int_{-\infty}^t|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|\|
\psi(s)f(s)\|ds \\
&\leqslant\int_{-\infty}^te^{-\alpha(t-s)}\|\psi(s)f(s)\|ds
 \leqslant c(1-e^{-\alpha})^{-1}
\end{align*}
 and
 \begin{align*}
\|\int_t^0\psi(t)Y(t)P_2Y^{-1}(s)f(s)ds\|
&\leqslant\int_t^0 e^{-\beta(s-t)}\|\psi(s)f(s)\|ds \\
&\leqslant\int_t^\infty  e^{-\beta(s-t)}\|\psi(s)f(s)\|ds
\leqslant c(1-e^{-\beta})^{-1}.
\end{align*}
It follows that the function
$$
\widetilde x(t)=\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)f(s)ds
 - \int_t^0\psi(t) Y(t)P_2Y^{-1}(s)f(s)ds
$$
is bounded on $\mathbb{R}_-$.
Hence the function
\begin{align*}
 x(t)&=\psi^{-1}(t)\widetilde x(t) \\
&=\int_{-\infty}^t\psi(t)Y(t)P_1Y^{-1}(s)f(s)ds
 - \int_t^0\psi(t) Y(t)P_2Y^{-1}(s)f(s)ds
\end{align*}
is $\psi$-bounded on $\mathbb{R}_-$. On the other hand
\begin{align*}
x'(t)&=A(t)(\int_{-\infty}^tY(t)P_1Y^{-1}(s)f(s)ds
 -\int_t^0Y(t)P_2Y^{-1}(s)f(s)ds)\\
&\quad +Y(t)P_1Y^{-1}(t)f(t)+Y(t)P_2Y^{-1}(t)f(t) \\
&= A(t)x(t)+f(t),
\end{align*}
it implies that $x(t)$ is a solution  of   \eqref{e1.1}.

In the proof of ``only if part'': The set
$$
\widetilde C_\psi=\{ x:\mathbb{R}_-\to\mathbb{R}^d : x
$$
is $\psi$-bounded and continuous on $\mathbb{R}_-\}$.
It is a Banach space with the norm
$\|x\|_{\widetilde C_\psi}=sup_{t\leqslant 0}\|\psi(t)x(t)\|$.
The first step: we show that  \eqref{e1.1} has a unique $\psi$-bounded
solution $x(t)$ with $x(0)\in\widetilde X_1  = P_1\mathbb{R}^d$
for each $f\in \widetilde C_\psi$  and
$\|x\|_{\widetilde C_\psi}\leqslant r\|f\|_{\widetilde C_\psi}$,
here $r$ is a positive constant  independent of $f$.

 The next  steps of  the proof are similar to the proof of
\cite[Theorem 3.3]{b1}, with the corresponding replacement
(for example replace $t\geqslant t_0\geqslant 0 $ by
$ 0\geqslant  t_0 \geqslant t$, $P_1$ by $-P_2, P_2$ by $-P_1$, $\infty$
by $-\infty$, $- \infty$ by $\infty$, \dots ).
\end{proof}

\begin{theorem} \label{thm3.2}
The equation \eqref{e1.1} has a unique $\psi$-bounded solution on
$\mathbb{R}$ for every $\psi$-integrally bounded  function $f$ on
$\mathbb{R}$ if and only if  \eqref{e1.2} has a $\psi$-exponential
 dichotomy on $\mathbb{R}$.
\end{theorem}

\begin{proof}
First, we prove the ``if'' part. By Lemma \ref{lem2.3} and in the same way as
in the proof of Theorem \ref{thm3.1},  the function
 $$
x(t)=\int_{-\infty}^t Y(t)P_1Y^{-1}(s)f(s)ds
 -\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds
$$
 is $\psi$-bounded and continuous  on $\mathbb{R}$.
Moreover,
\begin{align*}
x'(t)&=A(t)(\int_{-\infty}^t Y(t)P_1Y^{-1}(s)f(s)ds
-\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds)\\
&\quad + Y(t)P_1Y^{-1}(t)f(t)- Y(t)P_2Y^{-1}(t)f(t)\\
&= A(t)x(t)+f(t),
\end{align*}
it follows that $x(t)$ is a solution of  \eqref{e1.1}.

The uniqueness of the solution $x(t)$ result from  \eqref{e1.2} having
no nontrivial $\psi$-bounded solution on $\mathbb{R}$ (Remark \ref{rmk2.2}).
Suppose that $y$ is a $\psi$-bounded solution of  \eqref{e1.1} then
$x - y$ is a $\psi$-bounded solution of  \eqref{e1.2} on $\mathbb{R}$.
We conclude $x = y$ since $x - y$ is the trivial solution of \eqref{e1.2}.

We prove the ``only if ''part. Suppose that  \eqref{e1.1} has unique
$\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally bounded
 function $f$ on $\mathbb{R}$, we have to prove that \eqref{e1.1} has a
$\psi$-exponential dichotomy on $\mathbb{R}$. By
Lemma \ref{lem2.4}, Theorem \ref{thm1.4}
and Theorem \ref{thm3.1} we get  \eqref{e1.2} has a $\psi$-exponential
dichotomy on $\mathbb{R}_+$  with a pair of the supplementary
projections $P_1, P_2$ and has a $\psi$-exponential dichotomy
on  $\mathbb{R}_ -$.  with a pair of the supplementary projections
 $Q_1, Q_2$. Remark \ref{rmk2.2} follows that $P_1\mathbb{R}^d$  is the
subspace of $\mathbb{R}^d$ consisting of the values $x(0)$
of all $\psi$-bounded solutions $x(t)$ on $\mathbb{R}_+$
of  \eqref{e1.2} and $Q_2\mathbb{R}^d$ is the subspace of
$\mathbb{R}^d $ consisting of the values $x(0)$ of all $\psi$-bounded
solutions $x(t)$ on $\mathbb{R}_ -$ of  \eqref{e1.2}.
We are going to prove that
  \begin{equation} \label{e3.1}
\mathbb{R}^d =P_1\mathbb{R}^d  \oplus Q_2\mathbb{ R}^d .
\end{equation}
For each $u \in\mathbb{R}^d$, denote by $ x=x(t)$ the solution of
\eqref{e1.2}, $x(0) =u$. By Lemma \ref{lem2.4} we get $x =x_1 + x_2$,
where $x_1, x_2$ are the solutions of  \eqref{e1.2} corresponding
$\psi$-bounded on $\mathbb{R}_+$,$\mathbb{ R}_ -$.
It follows from Remark \ref{rmk2.2} that $x_1(0) \in P_1\mathbb{R}^d$ and
$x_2(0) \in Q_2\mathbb{R}^d$. It follows from $u = x_1(0) + x_2(0)$, that
  \begin{equation} \label{e3.2}
\mathbb{R}^d =P_1\mathbb{R}^d + Q_2\mathbb{ R}^d .
\end{equation}
By hypothesis  \eqref{e1.1}  with $f = 0$ has unique $\psi$-bounded
solution on $\mathbb{R}$ i.e.  \eqref{e1.2} have no nontrivial
$\psi$-bounded solution on $\mathbb{R}$. For any
$v \in P_1\mathbb{R}^d\cap Q_2\mathbb{R}^d$,
denote by $x(t)$  the solution of  \eqref{e1.2} such that $x(0) = v$.
Then $x(t)$  is the $\psi$-bounded solution of  \eqref{e1.2},
it implies that  $x(t)$ is the trivial solution. Hence $v = 0$. Consequently
  \begin{equation} \label{e3.3}
P_1\mathbb{R}^d\cap Q_2\mathbb{R}^d =0.
\end{equation}
    The relations \eqref{e3.2} and \eqref{e3.3} imply \eqref{e3.1}.
Now, we prove the existence of a pair supplementary projections,
for which  \eqref{e1.1} has a $\psi$-exponential dichotomy on $\mathbb{R}$.
Choose the projection $P$ of $\mathbb{R}^d$  such that
$ImP = P_1\mathbb{R}^d$ , $\ker P = Q_2\mathbb{R}^d$. By Lemma 2.1,
  \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}_+$,
 and  have a $\psi$-exponential dichotomy on $\mathbb{R}_ -$ with
the pair of the supplementary projections $P, I_d-P$.
From Remark \ref{rmk1.3} it
follows that  \eqref{e1.2} has a $\psi$-exponential dichotomy on
$\mathbb{R}$ with the pair of the supplementary projections $P, I_d-P$.
The proof  is complete.
\end{proof}

\begin{theorem} \label{thm3.3}
Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy on
$\mathbb{R}$. If
  \begin{equation} \label{e3.4}
\lim_{t\to\pm\infty}\int_t^{t+1}\|\psi(s)f(s)\|ds=0
\end{equation}
then the $\psi$-bounded 
solution of  \eqref{e1.1} is such that
   \begin{equation} \label{e3.5}
\lim_{t\to\pm\infty}\|\psi(t)x(t)\|=0.
\end{equation}
\end{theorem}

\begin{proof}
By Theorem \ref{thm3.2}, the unique solution of  \eqref{e1.1} is
$$
x(t) = \int_{-\infty}^tY(t)P_1Y^{-1}(s)f(s)ds
 -\int_t^\infty Y(t)P_2Y^{-1}(s)f(s)ds.
$$
\begin{equation} \label{e3.6}
\begin{aligned}
\|\psi(t)x(t)\|
&\leqslant \int_{-\infty}^t\|\psi(t)Y(t)P_1Y^{-1}(s)f(s)\|ds
  +\int_t^\infty \|\psi(t)Y(t)P_2Y^{-1}(s)f(s)\|ds \\
& \leqslant K\int_{-\infty}^t e^{-\alpha(t-s)}\|\psi(s)f(s)\|ds
  + L\int_t^\infty e^{-\beta(s-t)}\|\psi(s)f(s)\|ds \\
& \leqslant K_1\{\int_{-\infty}^t e^{-\alpha(t-s)}\|\psi(s)f(s)\|ds
  +\int_t^\infty e^{-\beta(s-t)}\|\psi(s)f(s)\|ds\},
\end{aligned}
\end{equation}
where $K1 = \max\{K,L\}$.
Denote by $\gamma = \min\{\alpha,\beta\}$. Under the hypothesis
\eqref{e3.4}, for a given $\varepsilon > 0$, there exists $T > 0$ such that
\[
\int_t^{t+1}\|\psi(s)f(s)\|ds < \frac{\varepsilon}{2K_1}(1-e^{-\gamma})
\quad \mbox{for } |t| >  T.
\]
Then from Lemma \ref{lem2.3} and inequality \eqref{e3.6} it follow that
\begin{align*}
\|\psi(t)x(t)\|&\leqslant K_1\frac{\varepsilon}{2K_1}(1-e^{-\gamma})
  [(1-e^{-\alpha})^{-1}+(1-e^{-\beta})^{-1}]\\
&\leqslant K_1\frac{\varepsilon}{2K_1}(1-e^{-\gamma})2(1-e^{-\gamma})^{-1}
=\varepsilon\quad \mbox{for all } |t| > T  ,
\end{align*}
this implies \eqref{e3.5}. The proof is complete.
\end{proof}

\begin{corollary} \label{coro3.4}
Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy on
$\mathbb{R}$. If
  \begin{equation} \label{e3.7}
\lim_{t\to\pm\infty}\|\psi(t)f(t)\|=0
\end{equation}
then the $\psi$-bounded
solution of  \eqref{e1.1} is such that
   \begin{equation} \label{e3.8}
\lim_{t\to\pm\infty}\|\psi(t)x(t)\|=0.
\end{equation}
\end{corollary}

\begin{proof} It is easy to see that \eqref{e3.7} implies \eqref{e3.4}
\end{proof}
Now, we consider the perturbed equation
\begin{equation} \label{e3.9}
{x'}(t)=[A(t)+B(t)]x(t)
\end{equation}
where $B(t)$ is a $d\times d$ continuous matrix function on
 $\mathbb{R}$. We have the following result.

\begin{theorem} \label{thm3.5}
Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$.
If $\delta=\sup_{t\in\mathbb{R}}\int_t^{t+1}|\psi(s)B(s)\psi^{-1}(s)|ds$ is
sufficiently small, then  \eqref{e3.9} has a $\psi$-exponential dichotomy
on $\mathbb{R}$.
\end{theorem}

\begin{proof}
By Theorem \ref{thm3.2} it suffices to show that the equation
\begin{equation}
{x'}(t)=[A(t)+B(t)]x(t)+f(t) \label{e3.10}
\end{equation}
has a unique $\psi$-bounded solution on $\mathbb{R}$ for every $\psi$-integrally
bounded $f$ function on $\mathbb{R}$. Denote by $G_\psi$ the set
$$
G_\psi=\{x:\mathbb{R}\to \mathbb{R}^d : x
\mbox{ is $\psi$-bounded and continuous on }\mathbb{R}\}.
$$
It is well-known that $G_\psi$ is a real Banach space with the norm
$$
\|x\|_{G_\psi}=\sup_{t\in R}\|\psi(t)x(t)\|.
$$
Consider the mapping $T: {G_{\psi}}\to {G_{\psi}}$ which is defined by
\begin{align*}
Tz(t)&=\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds \\
&\quad -\int_t^\infty Y(t)P_2Y^{-1}(s)[B(s)z(s)+f(s)]ds.
\end{align*}
It is easy verified that $Tz\in G_{\psi}$. More ever if
$z_1, z_2\in G_\psi$ then
\begin{align*}
&\|Tz_1-Tz_2\|_{G_\psi}\\
&\leqslant\int_{-\infty}^t|\psi(t)Y(t)P_1Y^{-1}(s)\psi^{-1}(s)|
|\psi(s)B(s)\psi^{-1}(s)|\|\psi(s)z_1(s)-\psi(s)z_2(s)\|ds\\
&\quad +\int_t^\infty|\psi(t)Y(t)P_2Y^{-1}(s)\psi^{-1}(s)|
 |\psi(s)B(s)\psi^{-1}(s)|\|\psi(s)z_1(s)-\psi(s)z_2(s)\|ds
\end{align*}
By Lemma \ref{lem2.3}, we have
\begin{align*}
\|Tz_1-Tz_2\|_{G_\psi}
&\leqslant K\|z_1-z_2\|_{G_\psi}\int_{-\infty}^t e^{-\alpha(t-s)}
 |\psi(s)B(s)\psi^{-1}(s)|ds \\
&\quad + L\|z_1-z_2\|_{G_\psi}\int_t^{\infty} e^{\beta(t-s)}|
 \psi(s)B(s)\psi^{-1}(s)|ds \\
&\leqslant \delta[K(1-e^{-\alpha})^{-1}+L(1-e^{-\beta})^{-1}]
  \|z_1-z_2\|_{G_\psi}
\end{align*}
Hence, by the contraction principle, if
$\delta[K(1-e^{-\alpha})^{-1}+L(1-e^{-\beta})^{-1}] < 1$,
then the mapping $T$ has a unique fixed point. Denoting this fixed point by $z$, we have
\begin{align*}
z(t)
&=\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)z(s)+f(s)]ds \\
&\quad - \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)z(s)+f(s)]ds.
\end{align*}
It follows that $z(t)$ is a solution on $\mathbb{R}$ of
\eqref{e3.10}.

 Now, we prove the uniqueness of this solution. Suppose that $x(t)$ is
 a arbitrary $\psi$-bounded solution on $\mathbb{R}$ of \eqref{e3.10}.
Consider the function

\begin{align*}
y(t)
&=x(t)-\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)x(s)+f(s)]ds \\
&\quad + \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)x(s)+f(s)]ds.
\end{align*}
It is easy to see that $y(t)$ is a $\psi$-bounded solution on $\mathbb{R}$
of  \eqref{e1.2}. Then from Theorem \ref{thm3.2} follows that $y(t)$ is the trivial solution. Then
\begin{align*}
x(t)
&=\int_{-\infty}^tY(t)P_1Y^{-1}(s)[B(s)x(s)+f(s)]ds\\
&\quad - \int_t^\infty Y(t)P_2 Y^{-1}(s)[B(s)x(s)+f(s)]ds.
\end{align*}
Hence $x(t)$ is the fixed point of mapping $T$. From the uniqueness of this point,
it follows that $x=z$. The proof is complete.
\end{proof}

\begin{corollary} \label{coro3.6}
Suppose that  \eqref{e1.2} has a $\psi$-exponential dichotomy on $\mathbb{R}$.
If $\delta=\sup_{t\in\mathbb{R}}|\psi(t)B(t)\psi^{-1}(t)|$ is
sufficiently small, then  \eqref{e3.9} has a $\psi$-exponential dichotomy
on $\mathbb{R}$.
\end{corollary}

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\end{document}